ASSIGNMENT 5 - SOLUTIONS
1. Find the centroid ofp
the solid bounded above by the sphere x2 + y 2 + z 2 = 4 and
below by the cone z = x2 + y 2 .
Solution. Let G be the given solid and denote its volume by
ZZZ
VG =
1 dV.
G
The coordinates of the centroid are given by
RRR
RRR
RRR
G x dV
G y dV
G z dV
x̄ =
, ȳ =
, z̄ =
.
VG
VG
VG
To compute each of these four integrals we use spherical coordinates. The solid G is
given by

 x = r sin φ cos θ
y = r sin φ sin θ

z=
r cos φ
where 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π4 (the limits of integration for φ are obtained,
by observing that the angle of the cone is π4 , see, for example, Assignment 4, Exercise
2
4). Note that ∂(x,y,z)
∂(r,φ,θ) = r sin φ. We first compute
ZZZ
Z 2π Z π Z 2
4
r2 sin φ dr dφ dθ
VG =
1 dV =
G
=
=
0
0
π
4
0
Z
Z π
8 2π 4
sin φ
dφ dθ =
sin φ dφ dθ
3 r=0
3 0
0
0
0
√
Z
Z
π
√
8 2π
8 2π
8π
2
4
[− cos φ]φ=0 dθ =
(1 −
) dθ =
(2 − 2).
3 0
3 0
2
3
Z
2π
Z
2
r3
Let us now compute
ZZZ
2π
Z
x̄VG =
π
4
Z
x dV =
G
π
4
Z
=
0
π
4
=
2 Z 2π
Z
0
Z
0
0
r3 sin2 φ cos θ dθ dr dφ
h
i2π
r3 sin2 φ sin θ
dr dφ = 0,
θ=0
0
ZZZ
Z
ȳVG =
2π
Z
y dV =
G
Z
=
π
4
0
Z
=
0
Z
0
2 Z 2π
0
π
4
Z
r3 sin2 φ cos θ dr dφ dθ
0
0
2
Z
0
2
Z
0
π
4
Z
2
r3 sin2 φ sin θ dr dφ dθ
0
r3 sin2 φ sin θ dθ dr dφ
0
2
h
i2π
r3 sin2 φ − cos θ
dr dφ = 0,
θ=0
0
1
2
ASSIGNMENT 5 - SOLUTIONS
Z
ZZZ
2π
Z
z dV =
z̄VG =
0
G
0
π
4
Z
2
r3 sin φ cos φ dr dφ dθ
0
Z 2π Z π
h r 4 i2
4 sin(2φ)
dφ dθ = 4
=
sin φ cos φ
dφ dθ
4 r=0
2
0
0
0
0
h cos(2φ) i π
4
= 2π.
= 2 · 2π · −
2
φ=0
Thus x̄ = 0, ȳ = 0, and
2π
3
√ =
√ .
z̄ = 8π
4(2 − 2)
3 (2 − 2)
Z
2π
Z
π
4
2. A fair die is rolled repeatedly five times. What is the probability that 1 turns up
exactly three times? What is the probability that 1 turns up at least three times?
Solution. Let X be the random variable representing the number of times 1 turns
up in the five rolls. Note that X has binomial distribution B(n; p), with n = 5(number
of trials) and p = 16 (the probability of ”success” in one trial). Then q = 1 − p = 65 ,
and the probability that 1 turns up exactly three times is
n 3 n−3
5
P (X = 3) =
p q
=
(1/6)3 (5/6)2 = 0.03215.
3
3
The probability that 1 turns up at least three times is
P (X ≥ 3) = P (X = 3) + P (X = 4) + P (X = 5)
5
5
5
4
3
2
(1/6) (5/6) +
(1/6)5 (5/6)0
(1/6) (5/6) +
=
4
5
3
= 0.0354938.
3. If the probability of producing a defective light bulb is p = 0.005, what is the
probability that a lot of 2000 light bulbs will contain more than 6 defectives?
Solution. Let X be the random variable that represents the number of defective
lightbulbs contained by the lot. X has a binomial distribution with n = 2000 and
p = 0.005, but since p is very small and n is large, we can approximate it by the
Poisson distribution with λ = np = 10. Hence
P (X ≤ 6) = P (X = 0) + P (X = 1) + P (X = 2)
+P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)
0
10
101 −10 102 −10 103 −10 104 −10 105 −10
=
· e−10 +
·e
+
·e
+
·e
+
·e
+
·e
0!
1!
2!
3!
4!
5!
≈ 0.13 = 13 %
so that
P (X > 6) = 1 − P (X ≤ 6) ≈ 87 %
4. If the lifetime X of a certain kind of automobile battery is normally distributed
with a mean of 4 yr and a variance of 1 yr, and the manufacturer wishes to guarantee
the battery for 3 yr, what percentage of the batteries is he expected to replace under
the guarantee?
Solution. P (X ≤ 3) = Φ( 3−4
1 ) = Φ(−1) = 0.1587 = 15.87 %
ASSIGNMENT 5 - SOLUTIONS
3
5. Let X be a random variable with binomial distribution B(n; p). What is the
probability that X takes an even value?
Solution. If k is an integer with 0 ≤ k ≤ n, the probability that X takes the value
k is
n k n−k
P (X = k) =
p q
,
k
where q = 1 − p. Hence X takes an even value with probability
X
n k n−k
n 0 n
n 2 n−2
n 4 n−4
p q
=
p q +
p q
+
p q
+ ...
P (X is even) =
k
0
2
4
k even
0≤k≤n
Consider now the binomial expansions
n X
n k n−k
n
(p + q) =
p q
k
k=0
n 0 n
n 1 n−1
n 2 n−2
n 3 n−3
n n n−n
=
p q +
p q
+
p q
+
p q
+ ... +
p q
,
0
1
2
3
n
n X
n
n
n
(q − p) = (−p + q) =
(−p)k q n−k
k
k=0
n 0 n
n 1 n−1
n 2 n−2
n 3 n−3
n
=
p q −
p q
+
p q
−
p q
+ ... +
(−p)n q n−n .
0
1
2
3
n
Note that when adding the two sums above, the ”odd” terms cancel, and we obtain
twice the sum of the ”even” terms, more precisely
n 0 n
n 2 n−2
n 4 n−4
n
n
(p + q) + (q − p) = 2
p q +2
p q
+2
p q
+ ...
0
2
4
X
n k n−k
= 2
p q
k
k even
0≤k≤n
= 2P (X is even).
Use now the fact that p + q = 1, to obtain
1
1
P (X is even) = [(p + q)n + (q − p)n ] = [1 + (1 − 2p)n ].
2
2